Exercise 24.8

Solution for $(a)$. Yes. Let $X=\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$, $x,y\in X$. Since each $X_{\alpha }$ is path connected, we have $f_{\alpha }:[0,1]\to X_{\alpha }$ continuous such that $f_{\alpha }(0)=x_{\alpha },f_{\alpha }(1)=y_{\alpha }$, where we assume the closed interval is $[0,1]$ after composition with multiplication and addition, which are continuous operations. Thus we have the function $f=(f_{\alpha })$, which is continuous by Theorem $19.6$, with $f(0)=x,f(1)=y$, and so $X$ is path-connected. □

Solution for $(b)$. No, since $\overline{S}$ in Example 24.7 is not path-connected while $S$ is: it is the image of the continuous map $x\mapsto (x,\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x))$ from ${ℝ}_{>0}$ to ${ℝ}^2$. □

Solution for $(c)$. Yes. For, let $x,y\in f(X)$, and choose $x_0\in f^{-1}(x),y_0\in f^{-1}(y)$. Then, there exists continuous $g:[a,b]\to X$ such that $g(a)=x_0,g(b)=y_0$, and so its composition $f\circ g:[a,b]\to Y$ is continuous with $(f\circ g)(a)=x,(f\circ g)(b)=y$. □

Solution for $(d)$. Yes. Let $x,y\in \bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ and $p\in \bigcap <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$. Then, there exists a continuous map $f:[a,b]\to \bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ with $f(a)=x,f(b)=p$, and similarly $g:[b,c]\to \bigcup <!--\; FUNCTION\; APPLICATION\; -->A_{\alpha }$ with $f(b)=p,f(c)=y$, since $a,p\in A_{\alpha }$ for some $\alpha$ and similarly for $y$ (we are free to have $\mathrm{Dom}<!--\; FUNCTION\; APPLICATION\; -->g=[b,c]$ by composition with multiplication and addition, which are continuous). Then, by the pasting lemma (Theorem 18.3) since $f,g$ are continuous and $f(b)=g(b)$, we see that $h=f$ on $[a,b]$ and $h=g$ on $[b,c]$ is a continuous map such that $h(a)=x,h(c)=y$. □